Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$ 
I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.

As an instance, I will justify an example inequality just to show you the intended strategy. Consider the inequality
$$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5}{x}.$$ For example, $2x^2 - 7$ can be written as $x^2 + (x^2 - 7)$. Because $x^2 - 7$ is a positive value for all $x < -\sqrt{7}$, removing it from the denominator makes the absolute value of the fraction greater. Also note that when $x < -\frac{21}{10}$, the numerator $|5x + 21| < 5|x|$, and this happens for all $x < -\sqrt{7}$. So $$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5|x|}{x^2} = \frac{5}{x}$$ as long as $x < -\sqrt{7}$.
 A: Hint:
Set
$$
f(x)=\frac{5x^3 + 3x^2 + x}{x^3 - x^2 - 1}.
$$
Then you have
$$
\lim_{x\to+\infty}f(x)=5\tag{1}
$$
Now use the limit definition to show that $f(x)<10$ for sufficiently large $x$. So you get
$$
\frac{f(x)}{x}<\frac{10}{x}
$$
for sufficiently large $x$.

Notes: The limit (1) implies that
$$
|f(x)-5|<1
$$
for sufficiently large $x$. So you have
$$
4<f(x)<6<10.
$$
A: Here is a simple and direct way using just some rough estimates:
For $x>2$ you have surely
$$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} \stackrel{x>2}{<} \frac{5x^2 + 3x^2 + x^2}{x^3 - (x^2 +1)} = \frac{9x^2}{x^3 - (x^2 +1)}$$
Since you want to have a $10$ in the numerator, note that
$$\frac 1{10}x^3 > x^2+1 \Leftrightarrow x^2(\frac x{10}-1)>1$$
which is surely satisfied for $x>20$. Hence,
$$\frac{9x^2}{x^3 - (x^2 +1)} \stackrel{x>20}{<} \frac{9x^2}{x^3 - \frac 1{10}x^3} = \frac{10}x$$
Done.
A: mrsamy's method is quite elegant, though it says nothing about how large $x\gg 1$ should be.
If you need a thighter inequality, the best way is to develop
$$f(x)=(5x^2+3x+1)x-10(x^3-x^2-1)$$
Graph it and notice it is negative for some $x$ close to (and below) $3$
Then just calculate $f(u+3)=-5u^3-32u^2-56u-5<0\quad$ for $\ u\ge 0$
Therefore the inequality is valid as soon as $x\ge 3$ (the actual value is $\approx 2.9057$).
A: Hint for a more direct approach not involving limits (at least not explicitly): if $x$ is large enough, then you can conclude
\begin{align*}
3x & < (0.01) x^2; \\
1 & < (0.01) x^2; \\
x^2 & < (0.01) x^3; \mathrm{~and} \\
1 & < (0.01) x^3.
\end{align*}
For such $x$, you will have:
$$5x^2 + 3x + 1 < 5x^2 + (0.01) x^2 + (0.01) x^2 = (5.02) x^2$$
and
$$x^3 - x^2 - 1 > x^3 - 0.01 x^3 - 0.01 x^3 = (0.98) x^3.$$
Therefore, you will be able to conclude:
$$\frac{5x^2 + 3x + 1} {x^3 - x^2 - 1} < \frac{(5.02) x^2}{(0.98) x^3} = \frac{5.02}{0.98} \cdot \frac{1}{x} < \frac{10}{x}.$$
